MHT CET 2026 19th April Morning Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2026 19th April Morning Shift

MCQ (Single Correct Answer)

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In $\triangle ABC$, with usual notations, if the sides $a$, $b$ and $c$ are in the ratio $18 : 17 : 7$, then $\cot\dfrac{A}{2} : \cot\dfrac{B}{2} : \cot\dfrac{C}{2} =$

$4 : 3 : 14$

$14 : 4 : 3$

$3 : 4 : 14$

$4 : 14 : 3$

MHT CET 2026 19th April Morning Shift

MCQ (Single Correct Answer)

-0

If $A = [a_{ij}]_{3\times3}$, where $a_{ij} = \begin{cases} 1, & \text{if } i+j \text{ is even} \\ 0, & \text{if } i+j \text{ is odd} \end{cases}$, then $\text{adj}(A) = \ldots$ ..

$\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$

$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{bmatrix}$

$\begin{bmatrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$

$\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$

MHT CET 2026 19th April Morning Shift

MCQ (Single Correct Answer)

-0

The inverse of matrix $\begin{bmatrix} 1+pq & p & 0 \\ q & 1+pq & p \\ 0 & q & 1 \end{bmatrix}$ is ...

$\begin{bmatrix} 1+pq & p & 0 \\ q & 1+pq & p \\ 0 & q & 1 \end{bmatrix}$

$\begin{bmatrix} 1 & p & p^2 \\ q & 1+pq & p+p^2q \\ q^2 & q+pq^2 & 1+pq+p^2q^2 \end{bmatrix}$

$\begin{bmatrix} 1 & -p & p^2 \\ -q & 1+pq & -(p+p^2q) \\ q^2 & -(q+pq^2) & 1+pq+p^2q^2 \end{bmatrix}$

$\begin{bmatrix} 1 & -p & p^2 \\ -q & 1+pq & p+p^2q \\ q^2 & q+pq^2 & 1+pq+p^2q^2 \end{bmatrix}$